What is the definition of “irrational algebraic expression”?
$begingroup$
Can somebody please tell what is the definition of "irrational algebraic expression"? Would this be analogous to the situation of irrational numbers?
What are some examples?
algebra-precalculus
edited Dec 27 '18 at 14:44
Blue
48.5k870154
$endgroup$
add a comment |
$begingroup$
Can somebody please tell what is the definition of "irrational algebraic expression"? Would this be analogous to the situation of irrational numbers?
What are some examples?
algebra-precalculus
edited Dec 27 '18 at 14:44
Blue
48.5k870154
$endgroup$
add a comment |
$begingroup$
Can somebody please tell what is the definition of "irrational algebraic expression"? Would this be analogous to the situation of irrational numbers?
What are some examples?
algebra-precalculus
edited Dec 27 '18 at 14:44
Blue
48.5k870154
$endgroup$
Can somebody please tell what is the definition of "irrational algebraic expression"? Would this be analogous to the situation of irrational numbers?
What are some examples?
algebra-precalculus
algebra-precalculus
edited Dec 27 '18 at 14:44
Blue
48.5k870154
edited Dec 27 '18 at 14:44
Blue
48.5k870154
edited Dec 27 '18 at 14:44
Blue
48.5k870154
edited Dec 27 '18 at 14:44
Blue
48.5k870154
edited Dec 27 '18 at 14:44
Blue
48.5k870154
48.5k870154
asked Dec 27 '18 at 14:43
user629353
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An irrational algebraic expression is an algebraic expression (Wikipedia: Algebraic expression) that contains a root sign or a power with broken rational exponent and is not a rational expression .
examples: $1-sqrt{3}$, $1-3^{frac{1}{2}}$
counterexamples: $1-sqrt{4}$, $1-4^{frac{1}{2}}$
answered Dec 27 '18 at 16:51
IV_IV_
1,345525
$endgroup$
$begingroup$
Thanks for answering
$endgroup$
– user629353
Dec 28 '18 at 3:11
add a comment |
$begingroup$
An algebraic number is a real number that is the root of a polynomial with integer coefficients. For instance, $sqrt 2$ is a root of the polynomial $x^2-2$, so $sqrt 2$ is an algebraic number. And $sqrt[5]{sqrt 3-1}$ is a root of the polynomial $(x^5+1)^2-3$, so it too is an algebraic number.
Not all algebraic numbers can be expressed as a combination of $n$th roots in this way $-$ in the jargon, not all polynomials are solvable by radicals. This is a key result of Galois theory.
Note that all rational numbers are algebraic, because $frac{p}{q}$ is a root of the polynomial $qx-p$.
answered Dec 27 '18 at 14:52
TonyKTonyK
42.6k355134
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053994%2fwhat-is-the-definition-of-irrational-algebraic-expression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An irrational algebraic expression is an algebraic expression (Wikipedia: Algebraic expression) that contains a root sign or a power with broken rational exponent and is not a rational expression .
examples: $1-sqrt{3}$, $1-3^{frac{1}{2}}$
counterexamples: $1-sqrt{4}$, $1-4^{frac{1}{2}}$
answered Dec 27 '18 at 16:51
IV_IV_
1,345525
$endgroup$
$begingroup$
Thanks for answering
$endgroup$
– user629353
Dec 28 '18 at 3:11
add a comment |
$begingroup$
An irrational algebraic expression is an algebraic expression (Wikipedia: Algebraic expression) that contains a root sign or a power with broken rational exponent and is not a rational expression .
examples: $1-sqrt{3}$, $1-3^{frac{1}{2}}$
counterexamples: $1-sqrt{4}$, $1-4^{frac{1}{2}}$
answered Dec 27 '18 at 16:51
IV_IV_
1,345525
$endgroup$
$begingroup$
Thanks for answering
$endgroup$
– user629353
Dec 28 '18 at 3:11
add a comment |
$begingroup$
An irrational algebraic expression is an algebraic expression (Wikipedia: Algebraic expression) that contains a root sign or a power with broken rational exponent and is not a rational expression .
examples: $1-sqrt{3}$, $1-3^{frac{1}{2}}$
counterexamples: $1-sqrt{4}$, $1-4^{frac{1}{2}}$
answered Dec 27 '18 at 16:51
IV_IV_
1,345525
$endgroup$
An irrational algebraic expression is an algebraic expression (Wikipedia: Algebraic expression) that contains a root sign or a power with broken rational exponent and is not a rational expression .
examples: $1-sqrt{3}$, $1-3^{frac{1}{2}}$
counterexamples: $1-sqrt{4}$, $1-4^{frac{1}{2}}$
answered Dec 27 '18 at 16:51
IV_IV_
1,345525
edited Dec 27 '18 at 16:59
answered Dec 27 '18 at 16:51
IV_IV_
1,345525
answered Dec 27 '18 at 16:51
IV_IV_
1,345525
answered Dec 27 '18 at 16:51
IV_IV_
1,345525
1,345525
$begingroup$
Thanks for answering
$endgroup$
– user629353
Dec 28 '18 at 3:11
add a comment |
$begingroup$
Thanks for answering
$endgroup$
– user629353
Dec 28 '18 at 3:11
$begingroup$
Thanks for answering
$endgroup$
– user629353
Dec 28 '18 at 3:11
$begingroup$
Thanks for answering
$endgroup$
– user629353
Dec 28 '18 at 3:11
add a comment |
$begingroup$
An algebraic number is a real number that is the root of a polynomial with integer coefficients. For instance, $sqrt 2$ is a root of the polynomial $x^2-2$, so $sqrt 2$ is an algebraic number. And $sqrt[5]{sqrt 3-1}$ is a root of the polynomial $(x^5+1)^2-3$, so it too is an algebraic number.
Not all algebraic numbers can be expressed as a combination of $n$th roots in this way $-$ in the jargon, not all polynomials are solvable by radicals. This is a key result of Galois theory.
Note that all rational numbers are algebraic, because $frac{p}{q}$ is a root of the polynomial $qx-p$.
answered Dec 27 '18 at 14:52
TonyKTonyK
42.6k355134
$endgroup$
add a comment |
$begingroup$
An algebraic number is a real number that is the root of a polynomial with integer coefficients. For instance, $sqrt 2$ is a root of the polynomial $x^2-2$, so $sqrt 2$ is an algebraic number. And $sqrt[5]{sqrt 3-1}$ is a root of the polynomial $(x^5+1)^2-3$, so it too is an algebraic number.
Not all algebraic numbers can be expressed as a combination of $n$th roots in this way $-$ in the jargon, not all polynomials are solvable by radicals. This is a key result of Galois theory.
Note that all rational numbers are algebraic, because $frac{p}{q}$ is a root of the polynomial $qx-p$.
answered Dec 27 '18 at 14:52
TonyKTonyK
42.6k355134
$endgroup$
add a comment |
$begingroup$
An algebraic number is a real number that is the root of a polynomial with integer coefficients. For instance, $sqrt 2$ is a root of the polynomial $x^2-2$, so $sqrt 2$ is an algebraic number. And $sqrt[5]{sqrt 3-1}$ is a root of the polynomial $(x^5+1)^2-3$, so it too is an algebraic number.
Not all algebraic numbers can be expressed as a combination of $n$th roots in this way $-$ in the jargon, not all polynomials are solvable by radicals. This is a key result of Galois theory.
Note that all rational numbers are algebraic, because $frac{p}{q}$ is a root of the polynomial $qx-p$.
answered Dec 27 '18 at 14:52
TonyKTonyK
42.6k355134
$endgroup$
An algebraic number is a real number that is the root of a polynomial with integer coefficients. For instance, $sqrt 2$ is a root of the polynomial $x^2-2$, so $sqrt 2$ is an algebraic number. And $sqrt[5]{sqrt 3-1}$ is a root of the polynomial $(x^5+1)^2-3$, so it too is an algebraic number.
Not all algebraic numbers can be expressed as a combination of $n$th roots in this way $-$ in the jargon, not all polynomials are solvable by radicals. This is a key result of Galois theory.
Note that all rational numbers are algebraic, because $frac{p}{q}$ is a root of the polynomial $qx-p$.
answered Dec 27 '18 at 14:52
TonyKTonyK
42.6k355134
edited Dec 27 '18 at 15:00
answered Dec 27 '18 at 14:52
TonyKTonyK
42.6k355134
answered Dec 27 '18 at 14:52
TonyKTonyK
42.6k355134
answered Dec 27 '18 at 14:52
TonyKTonyK
42.6k355134
42.6k355134
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053994%2fwhat-is-the-definition-of-irrational-algebraic-expression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
